Divergence Simplification/Identities

[feedmeister]

Quick question…

what does the following simplify to? Can it be written in any other way?

\nabla\bullet (a \bullet b)b

where a and b are vectors.

Thanks,

 

[issacnewton]

Quick question…

what does the following simplify to? Can it be written in any other way?

\nabla\bullet (a \bullet b)b

where a and b are vectors.

Thanks,

in general

\nabla (\varphi \mathbf{F})=(\nabla \varphi)\bullet\mathbf{F}+\varphi (\nabla \bullet \mathbf{F})

let \varphi = \mathbf{a}\bullet \mathbf{b}

and \mathbf{F}=\mathbf{b}

 

[feedmeister]

in general

\nabla (\varphi \mathbf{F})=(\nabla \varphi)\bullet\mathbf{F}+\varphi (\nabla \bullet \mathbf{F})

let \varphi = \mathbf{a}\bullet \mathbf{b}

and \mathbf{F}=\mathbf{b}

Thanks, but I didn't think that \nabla\bullet (\mathbf{a} \bullet \mathbf{b})\mathbf{b} was the same as \nabla (\varphi \mathbf{F})... there's still a \bullet between the \nabla and the rest of the statement.

Can you clarify?

 

[issacnewton]

in general

\nabla (\varphi \mathbf{F})=(\nabla \varphi)\bullet\mathbf{F}+\varphi (\nabla \bullet \mathbf{F})

little mishtake...above should be

\nabla \bullet (\varphi \mathbf{F})=(\nabla \varphi)\bullet\mathbf{F}+\varphi (\nabla \bullet \mathbf{F})

 

[feedmeister]

Thanks, IssacNewton.

When substituting in \mathbf{a}\bullet \mathbf{b} and \mathbf{b} into the equation, it looks like it'd simplifies further.. but it looks like it'd be ugly.

Any good way of simplifying it?

Thanks,

 

[issacnewton]

well you can use gradient of vector dot product to simplify

http://en.wikipedia.org/wiki/Vector_calculus_identities#Product_rule_for_the_gradient

but things WILL look ugly... you can probably simplify further using levi civita tensor

http://folk.uio.no/patricg/teaching/a112/levi-civita/

https://www.physicsforums.com/showthread.php?t=488903

google "levi civita tensor identities" for further help